Sphere winding, nonexact curvature and flat holonomy

Put $Q=x^2+y^2+z^2$. At every point with $Q>0$, [F5] gives a positive root $r_0$ of $s^2=Q$; the map $s\mapsto s^2$ on $(0,\mathrm{\infty })$ has derivative $2r_0\mathrm{\ne }0$, so [F6] gives a smooth local inverse near $Q$, which agrees with the unique positive root by [F5]. Composing this local inverse with the smooth function $Q$ proves that $r=\sqrt{Q}$ is smooth near each point of the punctured space. Thus the Cartesian coefficients of $F_b$ are smooth. On the dense open set $x^2+y^2>0$, write $xdy\wedge dz+ydz\wedge dx+zdx\wedge dy={\iota }_{x{\mathrm{\partial }}_x+y{\mathrm{\partial }}_y+z{\mathrm{\partial }}_z}(dx\wedge dy\wedge dz)$. The spherical-coordinate volume form is $r^2\sin \theta dr\wedge d\theta \wedge d\varphi$, and the radial vector is $r{\mathrm{\partial }}_r$, so contraction gives $r^3\sin \theta d\theta \wedge d\varphi$. Hence $F_b=b\sin \theta d\theta \wedge d\varphi$ and $d{F}_b=b\cos \theta d\theta \wedge d\theta \wedge d\varphi =0$. Since $d{F}_b$ is continuous by [F4], it vanishes on the closure, which is the whole punctured space. On the unit sphere use $G:(0,\pi )\times (0,2\pi )\to S^2$, $G(\theta ,\varphi )=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$. Its pullback of the outward area form ${\omega }_u(v,w)=\det (u,v,w)$ is $G^{\ast }\omega =\sin \theta d\theta \wedge d\varphi$, so $G$ preserves orientation. It is a diffeomorphism onto the sphere with the poles and the meridian $\varphi =0$ removed, extends smoothly to the closed rectangle, and its image has closure $S^2$. The domain is a bounded open Jordan rectangle, so [F15] gives ${\int }_{S^2}{F}_b=b{\int }_0^{2\pi }{\int }_0^{\pi }\sin \theta d\theta d\varphi =4\pi b$.

Let $\vartheta =-i{z}^{-1}dz$ be the invariant one-form on the fibre circle and let $\nu =-i{w}^{-1}dw$ be the base angular form in the statement. In a local angle $z=e^{i\theta }$, $\vartheta =d\theta$, so it evaluates to $1$ on the vertical generator; it is right invariant, as is ${\pi }^{\ast }\nu$. Thus $\omega =\vartheta +a{\pi }^{\ast }\nu$ on the trivial bundle is a principal connection: for each base tangent vector $X$, the unique vertical component $-a\nu (X){\mathrm{\partial }}_{\theta }$ makes $\omega$ vanish, so its kernel is a horizontal complement. Since the base circle has dimension one, $d\nu$ is a two-form on a one-dimensional manifold and is zero by [F4]. Along $w=e^{i\varphi }$, the explicit fibre curve $z(\varphi )=z_0{e}^{-ia\varphi }$ has $\dot{\theta }+a\dot{\varphi }=0$, so it is horizontal and ends at $z_0{e}^{-2\pi ia}$. Hence the holonomy is $e^{-2\pi ia}$, and $F=dA={\kappa }_{0}ad\nu =0$.

Suppose $F_b=dA$ for a global smooth one-form $A$ on $M$. Choose a nondegenerate convex tetrahedron $T\subset {\mathbb{R}}^3$ with the origin in its interior and with positively oriented ordered vertices $v_0,v_1,v_2,v_3$. For each face $j$, let ${\sigma }_j$ be its affine parametrization by the standard $2$-simplex with increasing remaining-vertex order and boundary coefficient $(-1{)}^j$. No face contains the origin, so the radial map $R(x)=x\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ is smooth on a neighborhood of every face and $R\circ {\sigma }_j$ is a smooth singular simplex. Shared edge restrictions are the same ordered affine edge maps with opposite boundary signs; hence $c={\sum }_{j=0}^3(-1{)}^{j}R\circ {\sigma }_j$ is a finite smooth singular cycle. For each outward face, write its plane as $n_j\cdot x=h_j$ with outward unit normal $n_j$ and $h_j>0$. Convexity and $0\in \mathrm{int}T$ imply every ray from the origin meets $\mathrm{\partial }T$ exactly once. Thus $R:\mathrm{\partial }T\to S^2$ is a homeomorphism; on each face interior it is a smooth diffeomorphism onto the corresponding open spherical patch, with inverse $u\mapsto h_{j}u\mathrm{/}(n_j\cdot u)$. These four patches are disjoint and their closures cover $S^2$, with overlaps only along the finitely many projected edges. If $v,w$ are tangent vectors to a face and oriented by its outward boundary orientation, the outward unit-sphere area form ${\omega }_u(a,b):=\det (u,a,b)$ satisfies

The radial projection therefore preserves the face orientation. Every point away from the projected edges has exactly one preimage, with positive local orientation, so its signed preimage count (the degree in this piecewise-smooth setting) is $+1$. For each odd $j$, precompose ${\sigma }_j$ with the affine transposition $\rho (u_1,u_2)=(u_2,u_1)$ of the standard triangle, and for each even $j$ leave it unchanged; call the resulting parametrizations ${\tilde{\sigma }}_j$. By the orientation convention in Integral of a form over a smooth singular simplex, ${\int }_{{\tilde{\sigma }}_j}{F}_b=(-1{)}^j{\int }_{{\sigma }_j}{F}_b$, and each $R\circ {\tilde{\sigma }}_j$ is orientation-preserving onto its face image. Their domains are bounded open Jordan triangles; they extend smoothly to their closures, and the whole sphere is covered by the closures of the four images. Applying [F15] to $F_b{\mathrm{\mid }}_{S^2}$ gives the exact reparameterization identity ${\int }_c{F}_b={\sum }_j{\int }_{{\tilde{\sigma }}_j}{F}_b={\int }_{S^2}{F}_b=4\pi b$; the projected edges have zero two-dimensional content and contribute no integral. By Stokes [F2], ${\int }_c{F}_b={\int }_{c}dA={\int }_{\mathrm{\partial }c}A=0$, contradicting $4\pi b\mathrm{\ne }0$. Thus no global potential exists when $b\mathrm{\ne }0$.

Since $r\ge \mathrm{\mid }z\mathrm{\mid }$ and $(r+z)+(r-z)=2r>0$, at least one of $r+z,r-z$ is positive, so $U_N,U_S$ cover $M$. On the north patch, $(1-\cos \theta )d\varphi =(xdy-ydx)\mathrm{/}(r(r+z))$; the denominator is nonzero there, so $A_N$ extends smoothly across the north axis. On the south patch, $(1+\cos \theta )d\varphi =(xdy-ydx)\mathrm{/}(r(r-z))$; the denominator is nonzero there, so $A_S$ extends smoothly across the south axis. Differentiating on the dense overlap gives $d{A}_N=d{A}_S=b\sin \theta d\theta \wedge d\varphi =F_b$. Both sides are smooth, so the identities extend by continuity to the respective full patches. Subtraction gives $A_S-A_N=-2bd\varphi$ on the overlap.

Let $u:M_0\to U(1)$ be any smooth gauge map and restrict it to $\gamma (\varphi )=(x_{\ast }^0,e^{i\varphi },y_{\ast },z_{\ast })$, with the other base coordinates fixed; write $v(\varphi )=u(\gamma (\varphi ))$. The chain rule [F10] and smooth group inversion [F3] make $v$ and $v^{-1}$ smooth. Since $\mathrm{\mid }v\mathrm{\mid }=1$, differentiating $\bar{v}v=1$ gives $\overline{v^{-1}{v}^{\mathrm{\prime }}}+v^{-1}{v}^{\mathrm{\prime }}=0$, so $q(\varphi ):=-i{v}^{-1}{v}^{\mathrm{\prime }}$ is real and smooth. It is continuous on $[0,2\pi ]$, hence integrable by [F7]. By [F12] and [F13], choose ${\chi }_0\in [0,2\pi )$ with $e^{i{\chi }_0}=v(0)$, and set $\chi (\varphi )={\chi }_0+{\int }_0^{\varphi }q(s)ds$; [F8] gives ${\chi }^{\mathrm{\prime }}=q$. By [F10] and [F14], $h=e^{i\chi }$ satisfies $h^{\mathrm{\prime }}=iqh$, while the definition of $q$ gives $v^{\mathrm{\prime }}=iqv$. Thus $(v{h}^{-1}{)}^{\mathrm{\prime }}=0$; applying [F9] to its real and imaginary parts, whose zero derivatives are integrable by [F7], yields $v{h}^{-1}=v(0)h(0{)}^{-1}=1$, so $v=e^{i\chi }$ on the full interval. Any other smooth phase lift $\tilde{\chi }$ of $v$ has the same derivative as $\chi$ by [F10] and [F14]; applying [F9] to $\tilde{\chi }-\chi$ shows the difference is constant. Since $v(2\pi )=v(0)$, $e^{i(\chi (2\pi )-\chi (0))}=1$, and [F11] gives $\chi (2\pi )-\chi (0)=2\pi n$ for some integer $n$, independently of the phase lift. Since this argument uses only a smooth circle-valued loop and its equal endpoint values, it proves the same phase-lift and winding claims for every smooth $g:S^1\to U(1)$. Hence ${\int }_{\gamma }(-i{u}^{-1}du)=2\pi n$.

Put $a_N=A_N\mathrm{/}{\kappa }_0$ and $a_S=A_S\mathrm{/}{\kappa }_0$. By [F3], the required transition convention is $a_S=a_N+d{f}_{NS}$ with $g_{NS}=e^{i{f}_{NS}}$. Here $f_{NS}=-2b\varphi \mathrm{/}{\kappa }_0$, so $g_{NS}=e^{-2ib\varphi \mathrm{/}{\kappa }_0}$. Replacing $\varphi$ by $\varphi +2\pi$ leaves this map unchanged exactly when $e^{-4\pi ib\mathrm{/}{\kappa }_0}=1$, equivalently $n:=2b\mathrm{/}{\kappa }_0\in \mathbb{Z}$. For this two-set cover set $g_{NN}=g_{SS}=1$ and $g_{SN}=g_{NS}^{-1}$; these maps obey the cocycle, and $a_N=a_S+\lambda (g_{SN})$ follows from $\lambda (g_{SN})=-\lambda (g_{NS})$, so both overlap directions satisfy [F3]. The degree of the smooth map $\varphi \mapsto e^{-in\varphi }$ is $-n$, while the flux quotient in step 1.1 is $n$. Under the set-size interface [A1], the conditional construction statement [L1] applies to these compatible countable-cover data and supplies the selected circle-bundle connection.

On any local angle arc $u=e^{i\psi }$, the gauge map sends $(x,e^{i\theta })$ to $(x,e^{i(\theta +\psi (x))})$, so [F10] and [F14] give $G_u^{\ast }\vartheta =\vartheta +d\psi =\vartheta -i{u}^{-1}du$. Its pullback adds $-i{u}^{-1}du$ to the base form of $\omega =\vartheta +a{\pi }^{\ast }\nu$, whose loop period is $2\pi (a+n)$ by step 2.3; consequently the holonomy remains $e^{-2\pi ia}$.

If $a\mathrm{\notin }\mathbb{Z}$, [F11] makes $e^{-2\pi ia}\mathrm{\ne }1$, so no gauge can make the connection trivial. If $a\in \mathbb{Z}$, the global map $u(x)=w^{-a}$ is smooth by [F3] because it is an integer power in the smooth circle group; on a local parameter $w=e^{i\varphi }$, [F10] and [F14] give $-i{u}^{-1}du=-ad\varphi =-a\nu$, cancelling the base form. Therefore gauge triviality is equivalent to $a\in \mathbb{Z}$.